Question

When 1.50 g of ba(s) is added to 100.00 g of water in a container open to the atmosphere, the reaction shown below occurs and the temperature of the resulting solution rises from 22.00°c to 33.10°c. if the specific heat of the solution is 4.18 j/(g ∙ °c), calculate δh for the reaction, as written. ba(s)+2h2o(l)→ba(oh)2(aq)+h2(g) δh=?

Answer 1

Answer:

The enthalpy of reaction = - 428.127 kJ

Explanation:

Enthalpy of a reaction is the amount of heat change during a reaction of one mole of a substance.

The heat released in the reaction is being absorbed by the solution.

The heat absorbed by solution = mass of solution X specific heat X change in temperature

Heat absorbed = (100+1.5) X 4.18 X (33.10-22) = 4709.397 Joules

This heat being released when 1.50g of Barium is added to water

The moles of Barium added =  mass / atomic mass = 1.5 / 137.33 = 0.011 moles

the heat released on adding 0.011 moles of barium = 4709.397 Joules

So heat released on adding 1 mole of barium = 428127 Joules = 428.127 kJ

The enthalpy of reaction = - 428.127 kJ

Answer 2

Given:

Mass of Ba = 1.50 g

Mass of H2O = 100.0 g

Initial temp T1 = 22 C

Final Temp T2 = 33.1 C

specific heat c = 4.18 J/g c

To determine:

The reaction enthalpy

Explanation:

The heat released during the reaction is:

q = - mc(T2-T1) = - (100+1.5) g *4.18 J/g C * (33.1-22) C = -4709.4 J

# moles of Ba = Mass of Ba/Atomic mass of Ba = 1.5 g/137 g.mol-1 = 0.0109 moles

ΔH = q/mole = - 4709.4 J/0.0109 moles = - 432 kJ/mol

Ans : The enthalpy change for the reaction is -432 kJ/mol